Let H be a an infinite dimensional and separable Hilbert space. Let C be a closed and bounded subset of H that is not compact. Does there always exist a closed and unbounded subset of H which is homeomorphic to C?
2 Answers
Yes. All we need is to construct a continuous on $H$ function $f$ that is unbounded on $C$. After that, $\{x,f(x)\}\subset H\times \mathbb C$ is closed, unbounded, and homeomorhic to $C$ (in the obvious way) and $H\times \mathbb C$ is isometric to $H$. Being closed and noncompact in an arbitrary separable metric space $X$ implies the existence of such a function. The simplest construction is to take a countable open cover $U_j$ of $X$ that contains no finite subcover of $C$ and to put $f(x)=\min f_j(x)$ where $f_j(x)=\frac j{\min(1,\operatorname{dist}(x,X\setminus U_j))}$. Separability has actually nothing to do with it but in the nonseparable case things become a bit more complicated.
Yes. Let $A$ be the set in question. We may assume that $0\notin A$ and moreover that $A$ is outside the unit ball centered at the origin.
Since $A$ is closed (in a complete space) and not compact, it contains an infinite set $\{p_i\}_{i\in\mathbb N}$ of points whose radial projections to the unit sphere are $\varepsilon$separated for some $\varepsilon>0$. For each $i$, consider a ray $R_i=\{tp_i:t\ge 1\}$. These rays are also $\varepsilon$separated away from one another. Let $U_i$ denote the $(\varepsilon/3)$neighborhood of $R_i$. It is easy to construct a homeomorphism $f_i:U_i\to U_i$ which maps $p_i$ to $i\cdot p_i$ and is the identity away from the $(\varepsilon/5)$neighborhood of $R_i$. The union of these maps $f_i$ and the identity map of $C\setminus \bigcup U_i$ is a homeomorphism from $C$ to itself that sends $A$ to an unbounded set.

$\begingroup$ A pointless nitpick: two $p_i$'s can be on the same ray (after all $p$ and $2p$ are $\varepsilon$separated...), so the separation condition for rays doesn't hold automatically. It is trivial to fix though: the radial projection of $A$ to the unit sphere is also not compact. $\endgroup$– fedjaSep 5, 2010 at 16:43

$\begingroup$ A minor point: maybe to get this family of $\epsilon$separated rays one should take the family $\{ p_i \}$ in the radial projection of $A$ on the unit sphere, rather than in $A$ itself (otherwise some $p_i$ could be multiple of each other). $\endgroup$ Sep 5, 2010 at 16:49

$\begingroup$ Thanks, I was thinking about the radial projections but wrote about the point. Fixed now. $\endgroup$ Sep 5, 2010 at 16:52

$\begingroup$ Call a point set "uncrowded" if it is denumerably infinite and if there exists a positive real number e such that every distinct pair of its points are at a distance apart not less than e. I was trying to prove the following two statements which would also imply a "yes" answer to my question. (!) Every closed and noncompact subset of H contains an "uncrowded" subset. (2) If A,B are "uncrowded" subsets of H then there exists a homeomorphism of H onto itself that carries A onto B. But your proofs, for which I thank you, are much better. $\endgroup$ Sep 6, 2010 at 15:02